1. O(n) additions and O(n²) comparisons
2. 1 + 2 + 3 + 4+…+ n < n + n + n + … n because 1 < n, 2 < n …

Therefore 1 + 2 + 3 + 4+…+ n < n ²

Let f(n) = 1 + 2 + 3 + 4+…+ n and g(n) = n ² .

Then there is a constant c = 1 and n₀ where f(n) < c * g(n) for n > n₀

Thus f(n) = O(g(n))